Ask question

Ask question

All categories

2 years ago

7

Which of the following states that an orbital can contain two electrons only if all other orbitals at that sublevel contain at l

east one electron?8

1 answer:

Vesnalui [34]2 years ago

5 0

Isn't that Hund's rule????

You might be interested in

Your theoretical yield is 81.2 grams, and your actual yield is 78.2 grams. What is the percent yield?

ioda

Answer:

66.2

Explanation:

3 0

2 years ago

A certain substance, X, has a triple-point temperature of 20°C at a pressure of 2.0 atm.

nikitadnepr [17]

Answer: The statement which could possibly not be true is C -" Liquid X can exist as a stable phase at 25°C, 1atm."

Explanation:

Triple point is the point where a substance co-exist as solid liquid and gas. At any point other than the triple point, the substance exist as a single phase substance.

As shown in the diagram, Liquid cannot exist as a stable phase at 1atm( below the the triple point pressure of 2atm) as the liquid can only exist beyond the pressure of triple point.

3 0

3 years ago

Which is the balanced equation for V205 + Cas - CaO + V S.?

barxatty [35]

Answer:c

Explanation:

7 0

3 years ago

How many grams N2F4 can be produced from 225 g F,?

zavuch27 [327]

Answer:

308 g

Explanation:

Data given:

mass of Fluorine (F₂) = 225 g

amount of N₂F₄ = ?

Solution:

First we look to the reaction in which Fluorine react with Nitrogen and make N₂F₄

Reaction:

2F₂ + N₂ -----------> N₂F₄

Now look at the reaction for mole ratio

2F₂ + N₂ -----------> N₂F₄

2 mole 1 mole

So it is 2:1 mole ratio of Fluorine to N₂F₄

As we Know

molar mass of F₂ = 2(19) = 38 g/mol

molar mass of N₂F₄ = 2(14) + 4(19) =

molar mass of N₂F₄ = 28 + 76 =104 g/mol

Now convert moles to gram

2F₂ + N₂ -----------> N₂F₄

2 mole (38 g/mol) 1 mole (104 g/mol)

76 g 104 g

So,

we come to know that 76 g of fluorine gives 104 g of N₂F₄ then how many grams of N₂F₄ will be produce by 225 grams of fluorine.

Apply unity formula

76 g of F₂ ≅ 104 g of N₂F₄

225 g of F₂ ≅ X of N₂F₄

Do cross multiplication

X of N₂F₄ = 104 g x 225 g / 76 g

X of N₂F₄ = 308 g

So,

308 g N₂F₄ can be produced from 225 g F₂

7 0

2 years ago

What are the similarities and differences for the properties of copper (II) sulphate and copper?

ivann1987 [24]

It is to do with the ionisation of the atom. Copper is a metal, so it will lose electrons. When reacted with a non-metal, it will form an ionic bond.

In copper (I) sulphate, the copper ions have a charge of +1, ie they have lost ONE electron each.

Copper (I) sulphate has the chemical formula Cu2SO4. Each ionic bond involves two Cu+1 ions and a sulphate ion (SO4.

In copper (II) sulphate, the copper ions have a charge of +2, ie they have lost TWO electrons each.

Copper (II) sulphate has the chemical formula CuSO4. Each ionic bond involves a single Cu+2 ion and a sulphate ion (SO4).

So, really, it’s down to the chemical structure and the ionisation of the atom. Apart from the chemistry, copper (I) sulphate a very obscure chemical. Although, after a bit of googling, I have managed to find some info and vendors, it appears that this chemical is rarely seen and doesn’t have many practical uses.

Copper (II) sulphate, on the other hand, is incredibly common. It’s in every school chemistry lab. If someone says “copper sulphate” they will be talking about this chemical, not copper (I) sulphate. In pure form, it is a boring white powder, but when hydrated, it takes on it’s better known blue colour, with blue crystals and blue solution.

Hope this helps.

3 0

2 years ago